Integral de $$$4^{- x}$$$

Halla $$$\int 4^{- x}\, dx$$$.

Sea $$$u=- x$$$.

Entonces $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - du$$$.

La integral puede reescribirse como

$${\color{red}{\int{4^{- x} d x}}} = {\color{red}{\int{\left(- 4^{u}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = 4^{u}$$$:

$${\color{red}{\int{\left(- 4^{u}\right)d u}}} = {\color{red}{\left(- \int{4^{u} d u}\right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=4$$$:

$$- {\color{red}{\int{4^{u} d u}}} = - {\color{red}{\frac{4^{u}}{\ln{\left(4 \right)}}}}$$

Recordemos que $$$u=- x$$$:

$$- \frac{4^{{\color{red}{u}}}}{\ln{\left(4 \right)}} = - \frac{4^{{\color{red}{\left(- x\right)}}}}{\ln{\left(4 \right)}}$$

Por lo tanto,

$$\int{4^{- x} d x} = - \frac{4^{- x}}{\ln{\left(4 \right)}}$$

Simplificar:

$$\int{4^{- x} d x} = - \frac{4^{- x}}{2 \ln{\left(2 \right)}}$$

Añade la constante de integración:

$$\int{4^{- x} d x} = - \frac{4^{- x}}{2 \ln{\left(2 \right)}}+C$$

$$$\int 4^{- x}\, dx = - \frac{4^{- x}}{2 \ln\left(2\right)} + C$$$A